soa exam P Prerequisite

Prerequisites for Exam P/1

PREREQUISITES FOR EXAM P/1 Much of probability and statistics is applied mathematics. The concepts that you need to know for this exam are built on a strong foundation in a select area in mathematics. This document will review many of the mathematical skills that you should be equipped with to ensure unobstructed progress when studying the material.

CONTENTS

1 Algebra ........................................................................................................................ 1

1.1 Quadratic Formula ......................................................................................................................................... 1 1.2 Laws of Exponents ......................................................................................................................................... 1 1.3 Laws of Logarithms ....................................................................................................................................... 2 1.4 Sum of Arithmetic Series ............................................................................................................................. 3 1.5 Sum of Geometric Series .............................................................................................................................. 3 1.6 Matrix Algebra ................................................................................................................................................. 5

2 Geometry .................................................................................................................... 6

2.1 Pythagorean Theorem .................................................................................................................................. 6 2.2 Areas .................................................................................................................................................................... 7

3 Differential Calculus ............................................................................................... 9

3.1 Basic Formulas and Rules ........................................................................................................................... 9 3.2 Chain Rule ........................................................................................................................................................ 10 3.3 Product Rule ................................................................................................................................................... 11 3.4 Quotient Rule ................................................................................................................................................. 11 3.5 Derivative of Absolute Value Functions............................................................................................... 12 3.6 Second Derivative ......................................................................................................................................... 13 3.7 Optimization ................................................................................................................................................... 14

4 Integral Calculus .................................................................................................... 16

4.1 Fundamental Theorem of Calculus ........................................................................................................ 16 4.2 Basic Formulas and Rules ......................................................................................................................... 17 4.3 Integration by Substitution ...................................................................................................................... 18 4.4 Integration by Parts ..................................................................................................................................... 20 4.5 Tabular Integration...................................................................................................................................... 22

5 Multivariate Calculus ........................................................................................... 25

5.1 Partial Derivatives ........................................................................................................................................ 25 5.2 Double Integration ....................................................................................................................................... 27

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1 Algebra 1.1 Quadratic Formula If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, then

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Example

𝑥2 − 𝑥 − 1 = 0

𝑎 = 1, 𝑏 = −1, 𝑐 = −1

𝑥 =1 ± √(−1)2 − 4(1)(−1)

2(1) ⇒ 𝒙𝟏 =𝟏 − √𝟓

𝟐, 𝒙𝟐 =

𝟏 + √𝟓𝟐

1.2 Laws of Exponents

𝑎� ⋅ 𝑎𝑛 = 𝑎�+𝑛

𝑎�

𝑎𝑛= 𝑎�−𝑛

(𝑎�)𝑛 = 𝑎�𝑛

𝑎�𝑛 = √𝑎�𝑛

𝑎0 = 1, 𝑎 ≠ 0

𝑎� ⋅ 𝑏� = (𝑎𝑏)�

𝑎�

𝑏�= (

𝑎𝑏)�

1𝑎𝑛

= 𝑎−𝑛

(𝑎𝑏)−𝑛

= (𝑏𝑎)𝑛

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1.3 Laws of Logarithms

𝑦 = 𝑒𝑥 ⇔ 𝑥 = ln(𝑦)

ln(𝑒𝑥) = 𝑒ln(𝑥) = 𝑥

ln(𝑥𝑛) = 𝑛 ln(𝑥)

ln(𝑥𝑦) = ln(𝑥) + ln(𝑦)

ln(𝑥 + 𝑦) ≠ ln(𝑥) + ln(𝑦)

ln (𝑥𝑦) = ln(𝑥) − ln(𝑦)

ln(𝑥 − 𝑦) ≠ ln(𝑥) − ln(𝑦)

Examples

𝑒ln(1

1−𝑡) =𝟏

𝟏 − 𝒕

𝑒−𝑥 ⋅ 𝑥 = 𝑒−𝑥 ⋅ 𝑒ln(𝑥) = 𝒆−𝒙+𝐥𝐧(𝒙)

𝑥√𝑥

+𝑒−2𝑥

𝑥4= 𝑥1−0.5 +

(𝑒−𝑥)2

(𝑥2)2 = 𝒙𝟎.𝟓 + (𝒆−𝒙

𝒙𝟐)𝟐

[𝑡𝑛

𝑡𝑛−2− 2 (

1𝑡)−1

+ 𝑡0]

12

= [𝑡𝑛−(𝑛−2) − 2𝑡 + 1]12

= (𝑡2 − 2𝑡 + 1)12

= [(𝑡 − 1)2]12

= 𝒕 − 𝟏 ln(𝑒−(𝑥2+𝑦2)) = −(𝒙𝟐 + 𝒚𝟐) ln(𝑥2 + 4𝑥 + 4) = ln(𝑥 + 2)2 = 𝟐 𝐥𝐧(𝒙 + 𝟐)

ln(𝑥2𝑦3

6𝑧) = ln(𝑥2𝑦3) − ln(6𝑧)

= [ln(𝑥2) + ln(𝑦3)] − [ln(6) + ln(𝑧)] = 𝟐 𝐥𝐧(𝒙) + 𝟑 𝐥𝐧(𝒚) − 𝐥𝐧(𝟔) − 𝐥𝐧(𝒛)

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1.4 Sum of Arithmetic Series The general formula is

𝑆𝑛 =𝑛2

(𝑇1 + 𝑇𝑛), where

𝑛 is the number of terms 𝑇1 is the first term of the series 𝑇𝑛 is the last term of the series

An expression for 𝑛 is

𝑛 =𝑇𝑛 − 𝑇1

𝑑+ 1

where 𝑑 is the common difference, defined as the difference between a term and the term prior to it. This is useful when we have the first and last terms but have to find the number of terms. Example

𝑆𝑛 = 1 + 5 + 9 + ⋯+ 41

𝑇1 = 1, 𝑇𝑛 = 41 , 𝑛 =41 − 1

4+ 1 = 11

𝑆𝑛 =112

(1 + 41) = 𝟐𝟑𝟏

1.5 Sum of Geometric Series Finite The standard formula is

𝑆𝑛 =𝑎(𝑟𝑛 − 1)𝑟 − 1

=𝑎(1 − 𝑟𝑛)

1 − 𝑟, where

𝑛 is the number of terms 𝑎 is the first term of the series 𝑟 is the common ratio, i.e., the ratio between a term and the term prior to it

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The next formula is derived from the previous formula but is easier to remember.

𝑆𝑛 =first term − first omitted term

1 − common ratio

The first omitted term is the term that would come after the last term if the series continued. Example

𝑆𝑛 = 1.02 + (1.02)2 + ⋯+ (1.02)20

𝑎 = 1.02, 𝑟 = 1.02, 𝑛 = 20 Using the standard formula,

𝑆𝑛 =1.02[(1.02)20 − 1]

1.02 − 1≈ 𝟐𝟒.𝟕𝟖𝟑𝟑

Using the alternative formula, the first omitted term is the term after (1.02)20, which is (1.02)21. Thus,

𝑆𝑛 =1.02 − (1.02)21

1 − 1.02≈ 𝟐𝟒.𝟕𝟖𝟑𝟑

Infinite If the common ratio is between −1 and 1 (non-inclusive), then an infinite geometric series converges to zero and its sum is finite, which equals

𝑆∞ =𝑎

1 − 𝑟

Example

𝑆∞ = 0.9 + 0.92 + 0.93 + ⋯

𝑎 = 0.9, 𝑟 = 0.9

𝑆∞ =0.9

1 − 0.9= 𝟗

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1.6 Matrix Algebra Suppose we have a two-by-two matrix, 𝐴, defined as follows:

𝐴 = [𝑎 𝑏𝑐 𝑑]

The determinant of 𝐴 is

|𝐴| = |𝑎 𝑏𝑐 𝑑| = 𝑎𝑑 − 𝑏𝑐

Example

The determinant of 𝐴 = [2 05 −2] is

|𝐴| = 2(−2) − 0(5) = −𝟒

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2 Geometry 2.1 Pythagorean Theorem

Figure 2.1

If 𝑎, 𝑏, and 𝑐 are the sides of a right triangle as shown in Figure 2.1, then

𝑎2 + 𝑏2 = 𝑐2 Example Find 𝑏 in Figure 2.2.

Figure 2.2

𝑏2 = 𝑐2 − 𝑎2

= (1 + 𝑥)2 − (1 − 𝑥)2 = 1 + 2𝑥 − 𝑥2 − 1 + 2𝑥 − 𝑥2 = 4𝑥

⇒ 𝑏 = √4𝑥 = 2√𝑥

c

a

b

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2.2 Areas

Diagram Formula for Area

Rectangle & square

Figure 2.3

𝐴 = 𝑏ℎ

Triangle

Figure 2.4

𝐴 =12𝑏ℎ

Parallelogram

Figure 2.5

𝐴 = 𝑏ℎ

Trapezoid

Figure 2.6

𝐴 = (average of 𝑎 and 𝑏) × ℎ

=12

(𝑎 + 𝑏)ℎ

b

h

b

h

b

h

b

h

a

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Circle

Figure 2.7

𝐴 = 𝜋𝑟2

Examples

Figure 2.8

Area of rectangle 𝐴 = 2 ⋅ 5 = 𝟏𝟎

Area of triangle 𝐴 =12⋅ 2 ⋅ 5 = 𝟓

Area of parallelogram 𝐴 = 1 ⋅ 3 = 𝟑

Area of trapezoid 𝐴 =12⋅ (1 + 3) ⋅ 2 = 𝟒

Area of circle 𝐴 = 𝜋 ⋅ 12 = 𝝅

r

80 1 2 3 4 5 6 7

6

1

2

3

4

5

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3 Differential Calculus 3.1 Basic Formulas and Rules

𝑑𝑑𝑥

𝑐 = 0

𝑑𝑑𝑥

𝑐𝑥 = 𝑐

𝑑𝑑𝑥

𝑥𝑐 = 𝑐 ⋅ 𝑥𝑐−1

𝑑𝑑𝑥

𝑒𝑥 = 𝑒𝑥

𝑑𝑑𝑥

𝑐𝑥 = 𝑐𝑥 ⋅ ln(𝑐) , 𝑐 > 0

𝑑𝑑𝑥

ln(𝑥) =1𝑥

, 𝑥 > 0

𝑑𝑑𝑥

[𝑓𝑋(𝑥) + 𝑔𝑋(𝑥)] = 𝑓𝑋�(𝑥) + 𝑔𝑋� (𝑥)

𝑑𝑑𝑥

𝑐 ⋅ 𝑓𝑋(𝑥) = 𝑐 ⋅𝑑𝑑𝑥

𝑓𝑋(𝑥)

𝑑𝑑𝑥

sin(𝑥) = cos(𝑥)

𝑑𝑑𝑥

cos(𝑥) = − sin(𝑥)

Note: 𝑐 is a constant. Examples

𝑑𝑑𝑥

(2𝑥 + 35) = 2 + 0 = 𝟐

𝑑𝑑𝑡

(𝑡 + 2)(2𝑡 − 3) =𝑑𝑑𝑡

(2𝑡2 + 𝑡 − 6) = 𝟒𝒕 + 𝟏

𝑑𝑑𝑥

(1𝑥3) =

𝑑𝑑𝑥

(𝑥−3) = −3𝑥−4 = −𝟑𝒙𝟒

𝑑𝑑𝑥

(2√𝑥 −1√𝑥

) =𝑑𝑑𝑥

(2𝑥0.5 − 𝑥−0.5) = 𝑥−0.5 + 0.5𝑥−1.5 =𝟏√𝒙

+𝟏

𝟐𝒙√𝒙

𝑑𝑑𝑡

ln(𝑡10) =𝑑𝑑𝑡

10 ln(t) = 10 ⋅1𝑡

=𝟏𝟎𝒕

𝑑𝑑𝑦

(𝑒𝑦 − 4𝑦) = 𝒆𝒚 − 𝟒𝒚 ⋅ 𝐥𝐧(𝟒)

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3.2 Chain Rule If 𝑓 and 𝑔 are differentiable functions where 𝑦 = 𝑓𝑋(𝑔𝑋(𝑥)), then

𝑑𝑦𝑑𝑥

= 𝑓𝑋�(𝑔𝑋(𝑥)) ⋅ 𝑔𝑋� (𝑥) This formula can be extended to include more than two differentiable functions. For three functions,

𝑦 = 𝑓𝑋 (𝑔𝑋(ℎ𝑋(𝑥))) 𝑑𝑦𝑑𝑥

= 𝑓𝑋� (𝑔𝑋(ℎ𝑋(𝑥))) ⋅ 𝑔𝑋� (ℎ𝑋(𝑥)) ⋅ ℎ𝑋� (𝑥)

Examples

𝑑𝑑𝑡

(4𝑡2 + 5𝑡)7 = 7(4𝑡2 + 5𝑡)6 ⋅𝑑𝑑𝑡

(4𝑡2 + 5𝑡)

= 𝟕(𝟒𝒕𝟐 + 𝟓𝒕)𝟔 ⋅ (𝟖𝒕 + 𝟓) 𝑑𝑑𝑥

(1

1 − 𝑥2) =

𝑑𝑑𝑥

(1 − 𝑥2)−1

= −(1 − 𝑥2)−2 ⋅𝑑𝑑𝑡

(1 − 𝑥2)

= −(1 − 𝑥2)−2 ⋅ (0 − 2𝑥)

=𝟐𝒙

(𝟏 − 𝒙𝟐)𝟐

𝑑𝑑𝑥

(1 − 𝑒−𝑥10) = 0 −

𝑑𝑑𝑥

(𝑒−𝑥10)

= −(𝑒−𝑥10) ⋅

𝑑𝑑𝑥

(−𝑥

10)

= −(𝑒−𝑥10) ⋅ (−

110)

=𝟏𝟏𝟎

𝒆−𝒙𝟏𝟎

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𝑑𝑑𝑥

ln(1 + 𝑥2) =1

1 + 𝑥2⋅𝑑𝑑𝑥

(1 + 𝑥2)

=1

1 + 𝑥2⋅ (0 + 2𝑥)

=𝟐𝒙

𝟏 + 𝒙𝟐

𝑑𝑑𝜃

(1 + sin𝜃)4 = 4(1 + sin𝜃)3 ⋅𝑑𝑑𝜃

(1 + sin𝜃)

= 4(1 + sin 𝜃)3 ⋅ (0 + cos 𝜃) = 𝟒 𝐜𝐨𝐬 𝜽 (𝟏 + 𝐬𝐢𝐧 𝜽)𝟑

3.3 Product Rule Let’s define 𝑢 = 𝑓𝑋(𝑥) and 𝑣 = 𝑔𝑋(𝑥), where 𝑓 and 𝑔 are differentiable functions. If 𝑦 = 𝑢 ⋅𝑣, then

𝑑𝑦𝑑𝑥

= 𝑢� ⋅ 𝑣 + 𝑢 ⋅ 𝑣� 3.4 Quotient Rule Let’s define 𝑢 = 𝑓𝑋(𝑥) and 𝑣 = 𝑔𝑋(𝑥), where 𝑓 and 𝑔 are differentiable functions. If 𝑦 = 𝑢

𝑣,

then

𝑑𝑦𝑑𝑥

=𝑣 ⋅ 𝑢� − 𝑢 ⋅ 𝑣�

𝑣2

Tip: It is not necessary to memorize the quotient rule, because 𝑢

𝑣 can be written as 𝑢 ⋅ 𝑣−1, which

is a product. Then, we can use the product rule instead.

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Example 1

𝑑𝑑𝑡

(3𝑡 + 1)2(𝑡2 + 2) = (𝑡2 + 2) ⋅𝑑𝑑𝑡

(3𝑡 + 1)2 + (3𝑡 + 1)2 ⋅𝑑𝑑𝑡

(𝑡2 + 2)

= (𝑡2 + 2) ⋅ 2(3𝑡 + 1) ⋅𝑑𝑑𝑡

(3𝑡 + 1) + (3𝑡 + 1)2 ⋅ 2𝑡

= (𝑡2 + 2) ⋅ 2(3𝑡 + 1) ⋅ 3 + (3𝑡 + 1)2 ⋅ 2𝑡 = 𝟔(𝟑𝒕 + 𝟏)(𝒕𝟐 + 𝟐) + 𝟐𝒕(𝟑𝒕 + 𝟏)𝟐

𝑑𝑑𝑥

[(4𝑥2 − 3)3

𝑒𝑥] =

𝑑𝑑𝑥

[(4𝑥2 − 3)3 ⋅ 𝑒−𝑥]

= 𝑒−𝑥 ⋅𝑑𝑑𝑥

(4𝑥2 − 3)3 + (4𝑥2 − 3)3 ⋅𝑑𝑑𝑥

𝑒−𝑥

= 𝑒−𝑥 ⋅ 3(4𝑥2 − 3)2 ⋅𝑑𝑑𝑥

(4𝑥2 − 3) + (4𝑥2 − 3)3 ⋅ 𝑒−𝑥 ⋅𝑑𝑑𝑥

(−𝑥)

= 𝑒−𝑥 ⋅ 3(4𝑥2 − 3)2 ⋅ 8𝑥 + (4𝑥2 − 3)3 ⋅ 𝑒−𝑥 ⋅ (−1) = 𝟐𝟒𝒙(𝟒𝒙𝟐 − 𝟑)𝟐𝒆−𝒙 − (𝟒𝒙𝟐 − 𝟑)𝟑𝒆−𝒙

3.5 Derivative of Absolute Value Functions We can write any absolute value function as follows:

|𝑓𝑋(𝑥)| = { 𝑓𝑋(𝑥), 𝑓𝑋(𝑥) ≥ 0

−𝑓𝑋(𝑥), 𝑓𝑋(𝑥) < 0

Then, the derivative of the absolute value function is the derivative of its individual components. 𝑑𝑑𝑥

|𝑓𝑋(𝑥)| = { 𝑓𝑋�(𝑥), 𝑓𝑋(𝑥) > 0

−𝑓𝑋�(𝑥), 𝑓𝑋(𝑥) < 0

The derivative may or may not be defined at 𝑓𝑋(𝑥) = 0.

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Example 1

|𝑥| = { 𝑥, 𝑥 ≥ 0−𝑥, 𝑥 < 0

𝑑𝑑𝑥

|𝑥| = { 𝑑𝑑𝑥

(𝑥), 𝑥 > 0

𝑑𝑑𝑥

(−𝑥), 𝑥 < 0= { 𝟏, 𝒙 > 0

−𝟏, 𝒙 < 0

The derivative is undefined at 𝑥 = 0.

Example 2

|𝑥2| = { 𝑥2, 𝑥2 ≥ 0

−𝑥2, 𝑥2 < 0

Because 𝑥2 cannot be negative, |𝑥2| = 𝑥2 for all 𝑥. Therefore, 𝑑𝑑𝑥

|𝑥2| = 𝟐𝒙, −∞ < 𝑥 < ∞ 3.6 Second Derivative The second derivative of a function 𝑓 is the derivative of the first derivative.

𝑑2

𝑑𝑥2𝑓𝑋(𝑥) =

𝑑𝑑𝑥

(𝑑𝑑𝑥

𝑓𝑋(𝑥))

Example

𝑦 =1

1 − 𝑥

𝑑𝑦𝑑𝑥

=𝑑𝑑𝑥

(1

1 − 𝑥)

=𝑑𝑑𝑥

(1 − 𝑥)−1

= −(1 − 𝑥)−2 ⋅𝑑𝑑𝑥

(1 − 𝑥)

=1

(1 − 𝑥)2

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𝑑2𝑦𝑑𝑥2

=𝑑𝑑𝑥

(𝑑𝑦𝑑𝑥

)

=𝑑𝑑𝑥

[1

(1 − 𝑥)2]

=𝑑𝑑𝑥

(1 − 𝑥)−2

= −2(1 − 𝑥)−3 ⋅𝑑𝑑𝑥

(1 − 𝑥)

=2

(1 − 𝑥)3

3.7 Optimization Derivatives can be used to find the maximum or minimum of a function, specifically the local max or min. We will illustrate this application using the following examples. Example 1 Find the local minimum of the function 𝑓𝑋(𝑥) = 𝑥2 − 2𝑥 − 5. Step 1: Take derivative of function.

𝑓𝑋�(𝑥) = 2𝑥 − 2 Step 2: Set derivative equal to zero, and solve for 𝑥.

𝑓𝑋�(𝑥) = 2𝑥 − 2 = 0 ⇒ 𝑥 = 1

If the question asks for the value of 𝑥 that minimizes 𝑓𝑋(𝑥), then the answer is 𝑥 = 1. To find the minimum value of 𝑓𝑋(𝑥), we continue with Step 3. Step 3: Substitute 𝑥 into function.

𝑓𝑋(1) = (1)2 − 2(1) − 5 = −6 The minimum value of 𝑓𝑋(𝑥) is −𝟔 .

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Step 4: Perform second derivative test (optional). If we know for sure that −6 is the minimum and not the maximum point, then we can skip this step. Otherwise, the second derivative test will tell us whether it is a minimum or maximum point. The second derivative of 𝑓𝑋(𝑥) is

𝑓𝑋��(𝑥) = 2 If 𝑓𝑋��(𝑥) is positive at the value of 𝑥 that we calculated, then the point is a minimum point. If 𝑓𝑋��(𝑥) is negative at the value of 𝑥 that we calculated, then the point is a maximum point. Because 𝑓𝑋��(1) = 2 > 0, (1,−6) is a minimum point, as shown in Figure 3.1.

Figure 3.1

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4 Integral Calculus 4.1 Fundamental Theorem of Calculus First Theorem The integral of a continuous function 𝑓 from 𝑎 to 𝑏 is

∫ 𝑓𝑋(𝑥) 𝑑𝑥𝑏

𝑎= 𝐹𝑋(𝑏) − 𝐹𝑋(𝑎), where 𝐹𝑋(𝑥) = ∫𝑓𝑋(𝑥) 𝑑𝑥

Second Theorem For a continuous function 𝑓,

𝑑𝑑𝑥

∫ 𝑓𝑋(𝑡) 𝑑𝑡𝑥

𝑎= 𝑓(𝑥), 𝑎 = constant

This is because

𝑑𝑑𝑥

∫ 𝑓𝑋(𝑡) 𝑑𝑡𝑥

𝑎=

𝑑𝑑𝑥

[𝐹𝑋(𝑥) − 𝐹𝑋(𝑎)]

= 𝑓𝑋(𝑥) − 0 = 𝑓𝑋(𝑥)

A more general version of the previous equation with functions 𝑢 = 𝑔𝑋(𝑥) and 𝑣 = ℎ𝑋(𝑥) in place of 𝑎 and 𝑥 is

𝑑𝑑𝑥

∫ 𝑓𝑋(𝑡) 𝑑𝑡𝑣

𝑢= 𝑓𝑋(𝑣) ⋅ 𝑣� − 𝑓𝑋(𝑢) ⋅ 𝑢�

Because the lower limit of integration is not a constant, its derivative is not zero. Furthermore, the derivatives of 𝑢 and 𝑣 are the result of the chain rule.

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4.2 Basic Formulas and Rules

∫𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶

∫𝑥𝑛 𝑑𝑥 =𝑥𝑛+1

𝑛 + 1+ 𝐶, 𝑛 ≠ −1

∫𝑥−1 𝑑𝑥 = ln(𝑥) + 𝐶

∫𝑏𝑎𝑥 𝑑𝑥 =1

𝑎 ln(𝑏)𝑏𝑎𝑥 + 𝐶

∫𝑒𝑎𝑥 𝑑𝑥 =1𝑎𝑒𝑎𝑥 + 𝐶

∫[𝑓𝑋(𝑥) + 𝑔𝑋(𝑥)] 𝑑𝑥 = ∫𝑓𝑋(𝑥) 𝑑𝑥 + ∫𝑔𝑋(𝑥) 𝑑𝑥

∫𝑎 ⋅ 𝑓𝑋(𝑥) 𝑑𝑥 = 𝑎 ⋅ ∫𝑓𝑋(𝑥) 𝑑𝑥

∫ sin(𝑥)𝑑𝑥 = − cos(𝑥) + 𝐶

∫ cos(𝑥)𝑑𝑥 = sin(𝑥) + 𝐶

Note: 𝑎, 𝑏, and 𝐶 are constants. Examples

∫ (5𝑥3 + 4) 𝑑𝑥1

0= [

5𝑥3+1

3 + 1+ 4𝑥]

0

1

= [54𝑥4 + 4𝑥]

0

1

= (54

+ 4) − (0 + 0)

=𝟐𝟏𝟒

∫ (1

2√𝑥− 𝑒−2.3𝑥) 𝑑𝑥

3

2= [

𝑥−0.5+1

2(−0.5 + 1) +1

2.3𝑒−2.3𝑥]

2

3

= [√𝑥 +1

2.3𝑒−2.3𝑥]

2

3

= (√3 +1

2.3𝑒−2.3(3)) − (√2 +

12.3

𝑒−2.3(2))

≈ 𝟎.𝟑𝟏𝟑𝟗

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∫25𝑡𝑑𝑡

10

1= [

25

ln(𝑡)]1

10

=25

[ln(10) − ln(1)]

≈ 𝟎.𝟗𝟐𝟏𝟎

∫ (𝑒−3𝑥 + 35𝑥) 𝑑𝑥0.1

0= [

𝑒−3𝑥

−3+

35𝑥

5 ln(3)]0

0.1

= (𝑒−3(0.1)

−3+

35(0.1)

5 ln(3)) − (1−3

+1

5 ln(3))

≈ 𝟎.𝟐𝟏𝟗𝟕 4.3 Integration by Substitution Integration by substitution is the reverse of the chain rule. We use integration by substitution if the integrand looks like it is the result of the chain rule. Example 1

∫ 2𝑥𝑒−𝑥2 𝑑𝑥2

1

Notice that 2𝑥 is the derivative of 𝑥2. So, we make the substitution 𝑢 = 𝑥2. Then, we convert the variable of integration from 𝑋 to 𝑈. The relationship between 𝑑𝑥 and 𝑑𝑢 comes from the derivative of 𝑢 = 𝑥2, i.e., 𝑑𝑢 = 2𝑥 𝑑𝑥. We convert the limits of integration by substituting them into 𝑢 = 𝑥2:

Lower limit: 𝑢 = 11 = 1 Upper limit: 𝑢 = 22 = 4

Therefore,

∫ 2𝑥𝑒−𝑥2 𝑑𝑥2

1= ∫ 𝑒−𝑢 𝑑𝑢

4

1

= [−𝑒−𝑢]14 = 𝑒−1 − 𝑒−4 ≈ 𝟎.𝟑𝟒𝟗𝟔

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Example 2

∫2𝑡

(𝑡 + 3)4 𝑑𝑡8

1

Identify the substitution 𝑢 = 𝑡 + 3, where 𝑡 = 𝑢 − 3. Then, we have 𝑑𝑢 = 𝑑𝑡. The new limits of integration are:

Lower limit: 𝑢 = 1 + 3 = 4 Upper limit: 𝑢 = 8 + 3 = 11

Therefore,

∫2𝑡

(𝑡 + 3)4 𝑑𝑡8

1= ∫

2(𝑢 − 3)𝑢4

𝑑𝑢11

4

= ∫ (2𝑢−3 − 6𝑢−4)𝑑𝑢11

4

= [2𝑢−2

−2+

6𝑢−3

3]4

11

= [−1𝑢2

+2𝑢3]4

11

= −1

112+

2113

+1

42−

243

≈ 𝟎.𝟎𝟐𝟒𝟓 Example 3

∫√𝑥

(1 + √𝑥)3 𝑑𝑥

9

4

Identify the substitution 𝑢 = 1 + √𝑥, where √𝑥 = 𝑢 − 1. Then, we have 𝑑𝑢 = 1

2√𝑥𝑑𝑥, so that 𝑑𝑥 = 2√𝑥 𝑑𝑢 = 2(𝑢 − 1) 𝑑𝑢.

The new limits of integration are:

Lower limit: 𝑢 = 1 + √4 = 3

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Upper limit: 𝑢 = 1 + √9 = 4 Therefore,

∫√𝑥

(1 + √𝑥)3 𝑑𝑥

9

4= ∫

𝑢 − 1𝑢3

⋅ 2(𝑢 − 1) 𝑑𝑢4

3

= 2∫𝑢2 − 2𝑢 + 1

𝑢3𝑑𝑢

4

3

= 2∫ (𝑢−1 − 2𝑢−2 + 𝑢−3)𝑑𝑢4

3

= 2 [ln(𝑢) + 2𝑢−1 −𝑢−2

2]3

4

= 2 (ln(4) +24−

12 ⋅ 42

) − 2 (ln(3) +23−

12 ⋅ 32

)

≈ 𝟎.𝟐𝟗𝟎𝟔 4.4 Integration by Parts Integration by parts is the reverse of the product rule.

𝑑𝑑𝑥

𝑢𝑣 = 𝑢𝑑𝑣𝑑𝑥

+ 𝑣𝑑𝑢𝑑𝑥

Integrating both sides of the equation,

∫𝑑(𝑢𝑣) = ∫𝑢 𝑑𝑣 + ∫𝑣 𝑑𝑢

𝑢𝑣 = ∫𝑢 𝑑𝑣 + ∫𝑣 𝑑𝑢

Then, the integration by parts formula is

∫𝑢 𝑑𝑣 = 𝑢𝑣 − ∫𝑣 𝑑𝑢

In general, integration by parts is applicable if the integrand is a product of two terms, one of which becomes zero when differentiated repeatedly.

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Example 1

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0

The first step is to identify 𝑢 and 𝑑𝑣, where 𝑢 is the term to be differentiated, and 𝑑𝑣 is the term to be integrated. In this example

𝑢 = 𝑥, 𝑑𝑣 = 𝑒−𝑥 𝑑𝑥 Then, we find 𝑑𝑢 and 𝑣.

𝑑𝑢 = 𝑑𝑥, 𝑣 = −𝑒−𝑥 Next, we apply the integration by parts formula as follows:

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0= [𝑥 ⋅ (−𝑒−𝑥)]0∞ − ∫ −𝑒−𝑥 𝑑𝑥

∞

0

= [−𝑥𝑒−𝑥]0∞ − [𝑒−𝑥]0∞ = [−𝑥𝑒−𝑥 − 𝑒−𝑥]0∞ = (0 − 0) − (0 − 1) = 𝟏

Example 2

∫ (3𝑥2 + 1)𝑒−𝑥10 𝑑𝑥

∞

0

The first step is to identify 𝑢 and 𝑑𝑣, where 𝑢 is the term to be differentiated, and 𝑑𝑣 is the term to be integrated. In this example

𝑢 = (3𝑥2 + 1), 𝑑𝑣 = 𝑒−𝑥10 𝑑𝑥

Then, we find 𝑑𝑢 and 𝑣.

𝑑𝑢 = 6𝑥 𝑑𝑥, 𝑣 = −10𝑒−𝑥10

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Next, we apply the integration by parts formula as follows:

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0= [(3𝑥2 + 1) ⋅ (−10𝑒−

𝑥10)]

0

∞− ∫ 6𝑥 ⋅ (−10𝑒−

𝑥10) 𝑑𝑥

∞

0

= [−10𝑒−𝑥10(3𝑥2 + 1)]

0

∞+ 60∫ 𝑥𝑒−

𝑥10 𝑑𝑥

∞

0

To evaluate the integral in blue, we use integration by parts a second time.

𝑢 = 𝑥, 𝑑𝑣 = 𝑒−𝑥10 𝑑𝑥

𝑑𝑢 = 𝑑𝑥, 𝑣 = −10𝑒−𝑥10

∫ 𝑥𝑒−𝑥10 𝑑𝑥

∞

0= [𝑥 ⋅ (−10𝑒−

𝑥10)]

0

∞− ∫ −10𝑒−

𝑥10 𝑑𝑥

∞

0

= [−10𝑥𝑒−𝑥10]

0

∞− [100𝑒−

𝑥10]

0

∞

= [−10𝑥𝑒−𝑥10 − 100𝑒−

𝑥10]

0

∞

= (0 − 0) − (0 − 100) = 100

In our final step,

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0= [(3𝑥2 + 1) ⋅ (−10𝑒−

𝑥10)]

0

∞+ 60(100)

= 0 − 1 ⋅ (−10) + 6,000 = 𝟔,𝟎𝟏𝟎

4.5 Tabular Integration Integration by parts can sometimes be messy to evaluate, which makes it prone to errors, especially when it needs to be done more than once. Tabular integration is an alternative to integration by parts that keeps our calculations organized. Besides, it takes less time and is less prone to errors. Tabular integration is best explained with examples. We will rework the two previous examples to show that integration by parts and tabular integration are equivalent.

Prerequisites for Exam P/1

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Example 1

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0

The first step is to choose the term that would be differentiated if were to do integration by parts, i.e., 𝑢. We write this term in the first row of the first column of the table (see Table 4.1). In this example, this term is 𝑥. Then, we differentiate 𝑥 repeatedly until we get zero, while writing down each derivative in a column. The derivative of 𝑥 is 1, and the derivative of 1 is 0. Next, we write the term that would be integrated if we were to do integration by parts in the first row of the second column. This term is 𝑒−𝑥. Then, we integrate 𝑒−𝑥 repeatedly until we have the same number of rows in both columns. Then, we draw diagonal lines from the first row of the first column to the second row of the second column, and from the second row of the first column to the third row of the second column. If we had more rows, we would continue drawing lines in this matter until we get to the last row of the second column. The next step is to label the lines with alternating signs, starting with “+” for the first line. We multiply the terms that are connected by the lines and then multiply by 1 if the line has a plus sign, or −1 if the line has a minus sign. We do the same for each pair of terms and add them to get the indefinite integral.

𝑥

𝑒−𝑥 + 1 −𝑒−𝑥 − 0 𝑒−𝑥

Table 4.1

Therefore,

∫ 𝑥𝑒−𝑥 𝑑𝑥∞

0= [+1 ⋅ 𝑥 ⋅ (−𝑒−𝑥) + (−1) ⋅ 1 ⋅ 𝑒−𝑥]0∞

= [−𝑥𝑒−𝑥 − 𝑒−𝑥]0∞ = (0 − 0) − (0 − 1) = 𝟏

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Example 2

∫ (3𝑥2 + 1)𝑒−𝑥10 𝑑𝑥

∞

0

The term to be differentiated is 3𝑥2 + 1. The term to be integrated is 𝑒−𝑥10.

3𝑥2+ 1

𝑒−𝑥10 +

6𝑥 −10𝑒−𝑥10 −

6 100𝑒−𝑥10 +

0 −1000𝑒−𝑥10

Table 4.2

Using Table 4.2,

∫ (3𝑥2 + 1)𝑒−𝑥10 𝑑𝑥

∞

0

= [+1 ⋅ (3𝑥2 + 1) ⋅ (−10𝑒−𝑥10) + (−1) ⋅ 6𝑥 ⋅ 100𝑒−

𝑥10 + (+1) ⋅ 6 ⋅ (−1000𝑒−

𝑥10)]

0

∞

= [−10(3𝑥2 + 1)𝑒−𝑥10 − 600𝑥𝑒−

𝑥10 − 6000𝑒−

𝑥10]

0

∞

= −0 − 0 − 0 + 10 + 0 + 6,000 = 𝟔,𝟎𝟏𝟎

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5 Multivariate Calculus 5.1 Partial Derivatives When taking the partial derivative of a multivariate function with respect to a variable, 𝑥, we treat all other variables as constants.

Partial derivative of 𝑓𝑋,�(𝑥,𝑦) with respect to 𝑥 𝜕𝜕𝑥

𝑓𝑋,�(𝑥, 𝑦)

Partial derivative of 𝑓𝑋,�(𝑥,𝑦) with respect to 𝑦 𝜕𝜕𝑦

𝑓𝑋,�(𝑥,𝑦)

Second order partial derivative of 𝑓𝑋,�(𝑥,𝑦) with respect to 𝑥

𝜕2

𝜕𝑥2𝑓𝑋,�(𝑥,𝑦) =

𝜕𝜕𝑥

(𝜕𝜕𝑥

𝑓𝑋,�(𝑥, 𝑦))

Second order partial derivative of 𝑓𝑋,�(𝑥,𝑦) with respect to 𝑦

𝜕2

𝜕𝑦2𝑓𝑋,�(𝑥,𝑦)

Second order partial derivative of 𝑓𝑋,�(𝑥,𝑦) with respect to 𝑥 and 𝑦

𝜕2

𝜕𝑥 𝜕𝑦𝑓𝑋,�(𝑥,𝑦) =

𝜕𝜕𝑥

(𝜕𝜕𝑦

𝑓𝑋,�(𝑥,𝑦))

𝜕2

𝜕𝑦 𝜕𝑥𝑓𝑋,�(𝑥,𝑦) =

𝜕𝜕𝑦

(𝜕𝜕𝑥

𝑓𝑋,�(𝑥,𝑦))

𝜕2

𝜕𝑥 𝜕𝑦𝑓𝑋,�(𝑥,𝑦) =

𝜕2

𝜕𝑦 𝜕𝑥𝑓𝑋,�(𝑥, 𝑦)

Example 1

𝑓𝑋,�(𝑥,𝑦) = 𝑥2𝑦

𝜕𝜕𝑥

(𝑥2𝑦) = 𝟐𝒙𝒚 Treat 𝑦 as constant.

𝜕𝜕𝑦

(𝑥2𝑦) = 𝒙𝟐 Treat 𝑥 as constant.

𝜕2

𝜕𝑥2(𝑥2𝑦) =

𝜕𝜕𝑥

(2𝑥𝑦) = 𝟐𝒚 Differentiate 𝜕

𝜕𝑥 with respect to 𝑥.

Treat 𝑦 as constant.

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𝜕2

𝜕𝑦2(𝑥2𝑦) =

𝜕𝜕𝑦

(𝑥2) = 𝟎 Differentiate 𝜕

𝜕𝑦 with respect to 𝑦.

Treat 𝑥 as constant.

𝜕2

𝜕𝑥 𝜕𝑦(𝑥2𝑦) =

𝜕𝜕𝑥

(𝑥2) = 𝟐𝒙 Differentiate 𝜕

𝜕𝑦 with respect to 𝑥.

Treat 𝑦 as constant. 𝜕2

𝜕𝑦 𝜕𝑥(𝑥2𝑦) =

𝜕𝜕𝑦

(2𝑥𝑦) = 𝟐𝒙 Differentiate 𝜕

𝜕𝑥 with respect to 𝑦.

Treat 𝑥 as constant. 𝜕2

𝜕𝑥 𝜕𝑦(𝑥2𝑦) =

𝜕2

𝜕𝑦 𝜕𝑥(𝑥2𝑦) Order of differentiation is reversible.

Example 2

𝑓𝑆,𝑇(𝑠, 𝑡) = 𝑒2𝑠2+𝑠𝑡+3𝑡2

𝜕𝜕𝑠

(𝑒2𝑠2+𝑠𝑡+3𝑡2) = 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅𝜕𝜕𝑠

(2𝑠2 + 𝑠𝑡 + 3𝑡2)

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (4𝑠 + 𝑡 + 0)

= (𝟒𝒔 + 𝒕)𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐 𝜕𝜕𝑡(𝑒2𝑠2+𝑠𝑡+3𝑡2) = 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅

𝜕𝜕𝑡

(2𝑠2 + 𝑠𝑡 + 3𝑡2)

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (0 + 𝑠 + 6𝑡)

= (𝒔 + 𝟔𝒕)𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐 𝜕2

𝜕𝑠2(𝑒2𝑠2+𝑠𝑡+3𝑡2) =

𝜕𝜕𝑠

[(4𝑠 + 𝑡)𝑒2𝑠2+𝑠𝑡+3𝑡2]

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅𝜕𝜕𝑠

(4𝑠 + 𝑡) + (4𝑠 + 𝑡) ⋅𝜕𝜕𝑠

(𝑒2𝑠2+𝑠𝑡+3𝑡2)

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (4 + 0) + (4𝑠 + 𝑡) ⋅ 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (4𝑠 + 𝑡 + 0)

= 𝟒𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐 + (𝟒𝒔 + 𝒕)𝟐𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐 𝜕2

𝜕𝑡2(𝑒2𝑠2+𝑠𝑡+3𝑡2) =

𝜕𝜕𝑡[(𝑠 + 6𝑡)𝑒2𝑠2+𝑠𝑡+3𝑡2]

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅𝜕𝜕𝑡

(𝑠 + 6𝑡) + (𝑠 + 6𝑡) ⋅𝜕𝜕𝑡(𝑒2𝑠2+𝑠𝑡+3𝑡2)

= 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (0 + 6) + (𝑠 + 6𝑡) ⋅ 𝑒2𝑠2+𝑠𝑡+3𝑡2 ⋅ (0 + 𝑠 + 6𝑡)

= 𝟔𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐 + (𝒔 + 𝟔𝒕)𝟐𝒆𝟐𝒔𝟐+𝒔𝒕+𝟑𝒕𝟐

Prerequisites for Exam P/1

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5.2 Double Integration We often use double integrals to calculate probabilities and moments. When evaluating double integrals, we always start with the inner integral and work our way out to the outer integral. There are two cases to consider: Case 1: Region of integration is rectangular. Case 2: Region of integration is not rectangular. Case 1 In the first case, double integrals are easier to evaluate. Besides, the order of integration is reversible without the need to redefine the limits of integration. In mathematical notation, this means

∫ ∫ 𝑓𝑋,�(𝑥,𝑦) 𝑑𝑦𝑏

𝑎𝑑𝑥

𝑑

𝑐= ∫ ∫ 𝑓𝑋,�(𝑥,𝑦) 𝑑𝑥

𝑑

𝑐𝑑𝑦

𝑏

𝑎

Figure 5.1 shows a rectangular region of integration.

Figure 5.1

x0 c d

y

a

b

Prerequisites for Exam P/1

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Example Integrate the following function over the shaded region in Figure 5.2.

𝑓𝑋,�(𝑥,𝑦) =𝑥𝑦

2500, 0 < 𝑥 < 10, 0 < 𝑦 < 10

Figure 5.2

If we chose 𝑦 as the variable for the inner integral, then the double integral is

∫ ∫𝑥𝑦

2500𝑑𝑦

6

2𝑑𝑥

7

1

Starting from the inside out, the inner integral is with respect to 𝑦, so we treat 𝑥 as a constant when performing the inner integral. Also, after performing the inner integral, we substitute the limits into 𝑦, not 𝑥, because the inner integral is with respect to 𝑦. Once we evaluate the inner integral, the double integral becomes a single integral that we evaluate in the usual way.

∫ ∫𝑥𝑦

2500𝑑𝑦

6

2𝑑𝑥

7

1= ∫ [

𝑥𝑦2

5000]𝑦=2

𝑦=6

𝑑𝑥7

1

= ∫𝑥

5000(62 − 22) 𝑑𝑥

7

1

x0 1 7

y

2

6

Prerequisites for Exam P/1

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=32

5000∫ 𝑥 𝑑𝑥7

1

=16

5000[𝑥2]17

=16

5000(72 − 12)

= 𝟎.𝟏𝟓𝟑𝟔 If we evaluate the same double integral in the reverse order of integration, we will get the same answer.

∫ ∫𝑥𝑦

2500𝑑𝑥

7

1𝑑𝑦

6

2= ∫ [

𝑥2𝑦5000

]𝑥=1

𝑥=7

𝑑𝑦6

2

= ∫𝑦

5000(72 − 12) 𝑑𝑦

7

1

=48

5000∫ 𝑦 𝑑𝑦6

2

=24

5000[𝑦2]26

=24

5000(62 − 22)

= 𝟎.𝟏𝟓𝟑𝟔 Case 2 In the second case, double integrals are trickier. They are reversible, but the limits of integration have to be redefined. We will demonstrate this with the following example. Example 1 Integrate the following function over the shaded region in Figure 5.3.

𝑓𝑋,�(𝑥,𝑦) = 𝑒−(𝑥+𝑦), 𝑥 > 0, 𝑦 > 0

Prerequisites for Exam P/1

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Figure 5.3

Step 1: Draw a diagram that satisfies all the constraints. In this example, the diagram has been given. However, on the exam, you often have to draw your own diagram. The constraints of the region are 𝑥 > 0, 𝑦 > 0, and 𝑥 + 𝑦 < 2. The region that satisfies these constraints is the shaded region in Figure 5.3. Note that Step 2 and Step 3 can be interchanged.

Step 2: Choose one variable for the outer integral. If we choose 𝑥, then the limits of the outer integration would be the minimum and maximum value of 𝑥 in the region. From the diagram, we see that the minimum value of 𝑥 is 0 and the maximum value of 𝑥 is 2. Thus, the lower limit is 0 and the upper limit is 2.

∫ (∫𝑓𝑋,�(𝑥,𝑦)𝑑𝑦)𝑑𝑥2

0

Likewise, if we choose 𝑦, then the limits of the outer integration would be the minimum and maximum value of 𝑦 in the region, which are 0 and 2, respectively.

∫ (∫𝑓𝑋,�(𝑥,𝑦)𝑑𝑥) 𝑑𝑦2

0

x0 1 2

y

1

2

y

x＋y= 2

Prerequisites for Exam P/1

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Both integrals will produce the same answer, but the variable that you choose for the outer integration will determine the order of integration; so you should choose the variable that will not require separating the region of integration, if possible. Step 3: Determine the limits of the inner integral. The third step is to define the rest of the integral. The variable for the inner integral will automatically be the variable not chosen for the outer integral. For example, if we chose 𝑥 for the outer integral, then the inner integral will be with respect to 𝑦. To determine the limits of integration when the integration is on 𝑦, draw a line parallel to the 𝑦-axis that cuts across the region of integration, as shown in Figure 5.4. The limits are the 𝑦-coordinates of the intersection points of the vertical line that was just drawn and the top and bottom bounds of the region.

Figure 5.4

In this example, the lower limit is 𝑦 = 0 because the vertical line intersects the lower bound at 0. Similarly, the upper limit is 𝑦 = 2 − 𝑥 because the vertical line intersects the upper bound at 2 − 𝑥. Since the vertical line can be drawn at any arbitrary location within the range of values of 𝑥, we always express the 𝑦-coordinates in terms of 𝑥, or the variable chosen for the outer integral. Therefore, our double integral is

x0 1 2

y

1

2

y

x＋y= 2

y = 2 − x

y = 0

Prerequisites for Exam P/1

Page 32 of 33

∫ (∫𝑓𝑋,�(𝑥,𝑦)𝑑𝑦)𝑑𝑥2

0= ∫ (∫ 𝑓𝑋,�(𝑥, 𝑦) 𝑑𝑦

2−𝑥

0)𝑑𝑥

2

0

On the other hand, if we chose 𝑦 for the outer integral, then the inner integral will be for the variable 𝑥. In this case, we draw a line parallel to the 𝑥-axis that cuts across the region, shown below in Figure 5.5.

Figure 5.5

We determine the limits of integration in the same way as before except the limits are now the 𝑥-coordinates of the intersection points of the horizontal line and the left and right bounds of the region, expressed in terms of 𝑦. In this example, the horizontal line intersects the left bound at 𝑥 = 0 and the right bound at 𝑥 = 2 − 𝑦. Therefore, our double integral is

∫ (∫𝑓𝑋,�(𝑥,𝑦)𝑑𝑥) 𝑑𝑦2

0= ∫ (∫ 𝑓𝑋,�(𝑥,𝑦) 𝑑𝑥

2−𝑦

0)𝑑𝑦

2

0

Step 4: Evaluate the double integral. The last step is to evaluate the double integral. There are 2 possible orders of integration, 𝑑𝑦 𝑑𝑥 or 𝑑𝑥 𝑑𝑦. Regardless, we always start with the inner integral and work our way out.

x0 1 2

y

1

2

y

x＋y= 2

x = 2 − yx = 0

Prerequisites for Exam P/1

Page 33 of 33

We will start with the 𝑑𝑦 𝑑𝑥 case.

∫ (∫ 𝑓𝑋,�(𝑥,𝑦) 𝑑𝑦2−𝑥

0)𝑑𝑥

2

0= ∫ (∫ 𝑒−(𝑥+𝑦) 𝑑𝑦

2−𝑥

0)𝑑𝑥

2

0

= ∫ [−𝑒−(𝑥+𝑦)]𝑦=0𝑦=2−𝑥

𝑑𝑥2

0

= ∫ [−𝑒−(𝑥+2−𝑥) − 𝑒−(𝑥+0)]𝑑𝑥2

0

= ∫ (−𝑒−2 + 𝑒−𝑥)𝑑𝑥2

0

= [−𝑒−2𝑥 − 𝑒−𝑥]02 = (−2𝑒−2 − 𝑒−2) − (0 − 𝑒−0) = 𝟏 − 𝟑𝒆−𝟐

Next, we will show that the 𝑑𝑥 𝑑𝑦 case produces the same answer.

∫ (∫ 𝑓𝑋,�(𝑥,𝑦) 𝑑𝑥2−𝑦

0) 𝑑𝑦

2

0= ∫ (∫ 𝑒−(𝑥+𝑦) 𝑑𝑥

2−𝑦

0) 𝑑𝑦

2

0

= ∫ [−𝑒−(𝑥+𝑦)]𝑥=0𝑥=2−𝑦

𝑑𝑦2

0

= ∫ [−𝑒−(2−𝑦+𝑦) − 𝑒−(0+𝑦)]𝑑𝑦2

0

= ∫ (−𝑒−2 + 𝑒−𝑦)𝑑𝑦2

0

= [−𝑒−2𝑦 + 𝑒−𝑦]02 = (−2𝑒−2 − 𝑒−2) − (0 − 𝑒−0) = 𝟏 − 𝟑𝒆−𝟐